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Transcript
Lecture 4: Gene interactions 1. 2. 3. 4. 5. 6. 7. Multiple alleles Codominance and incomplete dominance Lethal alleles Epistasis Complementation Penetrance and expressivity Chi-square (χ2 ) analysis Y/- : elimination of chlorophyll (no green color) y/y : keeps chlorophyll (green color) R/- : red, and r/r : yellow carotenoids c1 and c2: lighter shadows of red or yellow R/- ; y/y : brown (red plus green) Multiple alleles of the same gene Codominance Codominance P: IA/IA (blood group A) x IA / IB (blood group AB Î codominance ) F1: F2: IB/IB (blood group B) ¼ IA/IA 1 : ½ IA/IB 2 : ¼ IB/IB 1 The phenotype and the genotype ratios are the same (1:2:1) because heterozygotes show a distinct codominant phenotype. Codominance – both alleles contribute to phenotype in a heterozygote Incomplete dominance R/R r/r R/r “4 o’clock” plant: flowers get open in the afternoon and close by morning Incomplete dominance P: F1: F2: R/R (red) x r/r (white) R / r (pink) ¼ R/R 1 : ½R/r 2 : ¼ r/r 1 The phenotype and the genotype ratios are the same (1:2:1) and again heterozygotes show a distinct phenotype. But this time this is due to incomplete dominance – the dominant allele does not fully contribute to phenotype in a heterozygote (it is ‘weakened’ or ‘diluted’) Lethal alleles Parental yellow mice, NEVER true-breeding Lethal alleles Pleiotropy (mutiple effects) of AY This allels defines two different characters: yellow color (AY is dominant to A because AY/A are yellow) lack of viability (AY is recessive to A because AY/A are not dead) Epistasis, recessive Ability to make the pigment: C, full ability (full color) c, albino (no color) Distribution of the pigment: AY – no or inefficient distribution along the hair, yellow mice A – uneven, normal appearance mice (agouti) a – all over the hair, black mice Recessive epistasis: two recessive alleles of the epistatic gene (c/c), when present, affect (preven) expression of alleles of the hypostatic gene (A or a) 9:3:4 Epistasis, dominant Color of vegetables in squash Y: yellow; y: green W: no color (white); w: there is color (yellow or green) P: Y/Y; W/W (true breeding white) F1 x y/y; w/w (true breeding green) Y/y; W/w - all white F2 – how will it look like? Epistasis, dominant Dominant epistasis: a single dominant allele of the epistatic gene (W) is sufficient for the effect on Y or y Dominant epistasis: how it works Y/-; w/w : yellow y/y; w/w : green Y/-; W/- : white y/y; W/- : white Complementation Complementation: theory p/p; C/C x P/P; c/c Mutations in different genes p p P C P C p C P c c c Mutations in the same gene p1 p2 p2 p1 p1 p2 P/p; C/c no complementation complementation Complementation this is a mutant …but are the mutations in the same gene or in different genes? …and this is a mutant Complementation this is a mutant …and this is a mutant You cross the mutants and if you find that their progeny is wild type… and that means… Complementation this is a mutant …mutant alleles must belong to different genes! and this is a mutant Penetrance and expressivity The real situation may be very complicated Neurofibromatosis – highly variable expressivity Normal hands Brachydactyly (50-80% penetrance) Autosomal dominant: B/- brachydactyly, b/b normal fingers Chi-square analysis of test-cross results P: A/a; B/b x a/a; b/b (tester) F1 progeny Gametes AB ab Ab aB ab ab ab ab A/a; B/b a/a; b/b A/a; b/b a/a; B/b Observed # 140 135 110 115 Are these results consistent with 1:1:1:1 ratio, as predicted by Independent Assortment? 1. Choose the null hypothesis. In this case the hypothesis is the 1:1:1:1 ratio 2. Determine the Expected number of progeny in each class : E =125 3. Calculate chi-square value: (Observed – Expected)2 χ2 = 5.2 χ2 = ∑ Expected df = 3 4. Determine ‘degrees of freedom’ df = (number of classes) – 1 5. Use the df and χ2 values and Table 2.5 to determine P. This value is the probability of the observed difference between Observed and Expected being accidental (random or insignificant), that is due to chance alone 6. If P > 5% (0.05), this probability is considered high enough to accept that the differences are truly insignificant. Then we accept (cannot reject) the hypothesis 7. If P < 5% (0.05), this probability of the difference being insignificant is low; the difference are actually significant, and the ratio is different from the hypothesized. Reject the null hypothesis. 20% > P > 10% 5.2 Chi-square analysis of test-cross results P: A/a; B/b x a/a; b/b (tester) F1 progeny Gametes AB ab Ab aB ab ab ab ab A/a; B/b a/a; b/b A/a; b/b a/a; B/b Observed # 140 135 110 115 Are these results consistent with 1:1:1:1 ratio, as predicted by Independent Assortment? 1. Choose the null hypothesis. In this case the hypothesis is the 1:1:1:1 ratio 2. Determine the Expected number of progeny in each class : E =125 3. Calculate chi-square value: (Observed – Expected)2 χ2 = 5.2 χ2 = ∑ Expected df = 3 4. Determine ‘degrees of freedom’ df = (number of classes) – 1 5. Use the df and χ2 values and Table 2.5 to determine P. This value is the probability of the observed difference between Observed and Expected being accidental (random or insignificant), that is due to chance alone 6. If P > 5% (0.05), this probability is considered high enough to accept that the differences are truly insignificant. Then we accept (cannot reject) the hypothesis 7. If P < 5% (0.05), this probability of the difference being insignificant is low; the difference are actually significant, and the ratio is different from the hypothesized. Reject the null hypothesis.